2015-06-29 30 views
-2

爲什麼我的Modal不工作?使用POST方法是錯誤的嗎?當我加載頁面,並單擊鏈接忘記密碼和模態最終是顯示,然後當我輸入我的電子郵件沒有任何反應。有人能幫助我嗎?爲什麼我的Modal沒有通過html表單工作?

這是我的PHP腳本。

<?php 
require 'db.php'; 
$msg=''; 
if(isset($_POST['submit'])) 
{ 
$emailRet = mysqli_real_escape_string($connection, $_POST['emailRetrieve']); 

$sql2 = mysqli_query($connection, "SELECT Email_Address, Password FROM pawnshop WHERE Email_Address='".$emailRet."'"); 

while ($row = mysqli_fetch_array($sql2, MYSQL_ASSOC)) { 
    $emailRetrieve = $row['Email_Address']; 
    $passwordRetrieve = $row['Password']; 
} 

    $number = mysqli_num_rows($sql2); 
    echo $number; 

    if(mysqli_num_rows($sql2) < 1){ 
    require 'smtp/Send_Mail.php'; 

    $to = $emailRetrieve; 
    $subject = "PBMS Password Retrieval"; 
    $body ='Hi, This is your account information<br><br> Username:'.$emailRetrieve.'<br> Password:'.$passwordRetrieve.''; 

    Send_Mail($to,$subject,$body); 
    $msg='Check your email to get your password'; 
    echo 'send'; 
    } 
    else 
    { 
    $msg='Check your email and try again'; 
    } 

} 
?> 

這是模態

<html> 
    <head> 
     <link type='text/css' href='css/modal.css' rel='stylesheet' media='screen' /> 
     <script type='text/javascript' src='js/jqueryModal.js'></script> 
     <script type='text/javascript' src='js/jquery.simplemodal.js'></script> 
     <script type='text/javascript' src='js/modal.js'></script> 
    </head> 
    <body> 
    <form method="post"> 
    <div id='basic-modal'> 
       <center> 
       <a href='#' class='basic custom-font-reg'>Forgot password?</a><br> 
       <a href='Registration.php' class='custom-font-reg'>Register for an account?</a> 
       </center> 
       </div> 

       <hr class="hr-custom2"> 

       <div id="basic-modal-content"> 
       <div class="getPasswordWrapper"> 
       <hr class="PasswordHeaderColor"></hr> 
       <p class="enterEmailPasswordText">Enter your username to get your password</p> 
       <center> 
         <input type="text" name="emailRetrieve" class="getPassword" placeholder="Your Username\Email" required/> 
         <button type="submit" class="getPasswordButton">Send</button> 
         <span><?php echo $msg; ?></span> 
       </center> 
       </div> 
       </div> 
    </form> 
    </body> 
    </html> 

回答

2

首先,你不必與「提交」 name屬性的提交與$_POST['submit']一起去吧,這就是它失敗的原因,所以if(isset($_POST['submit'])){...}裏沒有任何東西會被執行。

所以做到這一點:

<button type="submit" class="getPasswordButton" name="submit">Send</button> 

或輸入:

<input type="submit" class="getPasswordButton" name="submit" value="Send"> 

另外,MYSQL_ASSOC必須MYSQLI_ASSOC。你不能混用MySQL的功能。

但是,看if(mysqli_num_rows($sql2) < 1){,可能需要使用>運算符進行更改。做< 1告訴MySQL它是否不存在; 瞭解

1

我的HTML代碼,你的表格必須有一個action屬性。它的值必須是處理表單的腳本文件的名稱。

<form action="your/php/script.php" method="post"> 

,或者,如果PHP腳本與形式相同的頁面上,你可以寫:

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> 
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